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I know very little about fields but I have been recently interested in sequences of algebraic structures of a certain type, with n elements (e.g. OEIS A000001 and relevant enumerations), and have not found any such count for fields. I came across the fact that fields cannot exist with cardinality (this is the correct term for the 'order' of a field analogous to the group-theoretic sense, yes?) that is not a power of a prime. This in and of itself doesn't really seem to be enough to prevent the enumeration of fields of cardinality n, since you can just adjust it to be fields of cardinality of the nth number in the sequence of prime powers, so I wonder if all finite fields of a given cardinality are isomorphic. That would indeed render such a count trivial.

So, can more than one field of a given (finite) cardinality exist? If so, what kinds of methods do there exist to enumerate them? I expect their growth rate, if it even exists, would be sub-linear since if I recall correctly abelian groups grow linearly, though I do in fact suspect that the growth rate is nonexistent (every prime-power order = 1). If this is in fact the case, what exactly is the constraining property? Or more precisely I suppose, how could one relax the field axioms to produce a similar sequence that has a growth rate (no matter how small)?

May Emerson
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    Given a natural number $n$, there are (up to isomorphism) either $0$ or $1$ fields with $n$ elements. The number of such fields is $1$ when and only when $n=p^k$ for some natural prime $p$ and some natural $k$. – José Carlos Santos Sep 03 '23 at 07:02
  • @JoséCarlosSantos That's a full answer. Why are you posting a full answer in the comment section? – Arthur Sep 03 '23 at 07:40
  • @Arthur Because I did not provide a proof and also because I strongly suspect that this question is a duplicate. – José Carlos Santos Sep 03 '23 at 07:50
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    See e.g. https://math.stackexchange.com/questions/633056/the-proof-that-a-finite-field-has-a-prime-power-order [every finite field has prime power order], https://math.stackexchange.com/questions/3960880/uniqueness-of-finite-fields-with-pn-elements [uniqueness of finite field of a given order] – leslie townes Sep 03 '23 at 07:59
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    You can try to find the number of finite-dimensional algebras over a finite field. For instance, how many commutative $2$-dimensional $\mathbb{F}_2$-algebras are there? How many non-commutative $2$-dimensional $\mathbb{F}_2$-algebras are there? – Gunnar Sveinsson Sep 03 '23 at 08:20
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    For what it's worth, if any set theorists in the crowd would like to give an answer for infinite cardinalities, I would be interested. A respectable amount of digging didn't turn anything up. – Eric Nathan Stucky Sep 03 '23 at 09:51
  • @Eric: there are continuum many countable fields, e.g. you can take $\mathbb{Q}$ and adjoin the square roots of any subset of the primes. I don't know what happens for larger cardinalities. – Qiaochu Yuan Sep 13 '23 at 19:01

1 Answers1

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Every finite field of a given cardinality is isomorphic, yes.

Here is what I think is the shortest proof "from scratch." First, every field has a prime subfield, namely the subfield generated by $1$. If $F$ is a finite field, the prime subfield must be $\mathbb{F}_p$ for some prime $p$, and $F$ is finite-dimensional as a vector space over $\mathbb{F}_p$ and so must have cardinality $p^n$ where $n = \dim_{\mathbb{F}_p} F$.

Next, the multiplicative group $F^{\times}$ has order $p^n - 1$, so by Lagrange's theorem every nonzero element satisfies $x^{p^n - 1} = 1$, and since there are $p^n - 1$ nonzero elements these must constitute exactly the roots of this polynomial. It follows that $F$ is isomorphic to the splitting field of $x^{p^n - 1} - 1$ in the algebraic closure $\overline{\mathbb{F}_p}$, which is unique given its cardinality $p^n$. (This argument can be replaced with a more explicit argument using the primitive element theorem: you write $F$ as $\mathbb{F}_p[\alpha]/f(\alpha)$ for an irreducible polynomial $f$ of degree $n$, then by the above argument $f(x) \mid x^{p^n - 1} - 1$, and since this is true for every irreducible polynomial of degree $n$ this shows that $F$ does not actually depend on the choice of $f$, since in fact it contains (and is generated by) a root of every other irreducible polynomial of degree $n$. There is also a third argument using the Frobenius map $x \mapsto x^p$.)

The only thing left to do from here is to show that for every prime power $p^n$ there actually exists a finite field of that order. This can be done using Mobius inversion to count the number of monic irreducible polynomials over $\mathbb{F}_p$ of degree $n$; the answer turns out to be a necklace polynomial

$$\frac{1}{n} \sum_{d \mid n} \mu(d) p^{\frac{n}{d}}$$

and this count is always positive; in fact it's at least $\frac{p^n - p^{n-1} - \dots - 1}{n}$ and the numerator is positive because $p \ge 2$.

Edit: Ah, actually there's a simpler argument. Once we're willing to grant that the algebraic closure $\overline{\mathbb{F}_p}$ exists, we can construct $\mathbb{F}_{p^n}$ as the set of roots of $x^{p^n} - x$ inside of it. This is the fixed subfield of the $n^{th}$ power of the Frobenius map $x \mapsto x^p$ so it is in particular a finite field, and it has order $p^n$ because we can check by taking the formal derivative (which is $-1$) that this polynomial is separable and therefore has no repeated roots.

As for your last question, you can generalize from fields to semisimple rings. Every semisimple commutative ring is (uniquely, up to permutation) a finite direct product of fields, so every finite semisimple commutative ring is a finite direct product of finite fields. Funnily enough this implies that the count of finite semisimple commutative rings is exactly the same as the count of finite abelian groups, which is A000688.

Qiaochu Yuan
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